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\author{林康益 2019211761}
\title{Hyperbolic functions}
\subtitle{the relation with hyperbola}
\institute{School of Mathematic and Statistics of CCNU}
\date{May 7, 2021}
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\begin{document}

\begin{frame}
	\titlepage
	\begin{figure}[htpb]
		\begin{center}
			\includegraphics[width=0.2\linewidth]{MSSCCMUlogo.pdf}
			\includegraphics[width=0.2\linewidth]{wechantqr.jpg}
		\end{center}
	\end{figure}
\end{frame}
\begin{frame}
\tableofcontents[sectionstyle=show,subsectionstyle=show/shaded/hide,subsubsectionstyle=show/shaded/hide]
\end{frame}

\section{Trigonometric function}



\begin{frame}[allowframebreaks]{Trigonometric function}
	\begin{columns}[c]
	\column{.45\textwidth} % Left column and width
	For The unit circle,
	\begin{tikzpicture}
		%定义坐标
		\coordinate (o) at (0,0);
		\coordinate (a) at (30:1.5cm);
		\coordinate (c) at (0:1.5cm);
		%画x和y轴坐标
		\draw[<->](2.1,0)--(o)--(0,2.1); \node[right] at (2.1,0){x};\node[above] at (0,2.1){y};
		\draw(-2.1,0)--(o)--(0,-2.1);
		%画刻度
	   %\node[below] at (1.6,0){1};\node[below] at (-1.6,0){-1};
	   %\node[left] at (0,1.6){1}; \node[left] at (0,-1.6){-1}; 
	   %画原点
	   \node[below] at (o){O};
		%画圆
	   \draw (o) circle (1.5cm);
	   \draw (o)--(30:1.5cm);
	   %画虚线
	   \draw[dashed] (1.299,0)--(1.299,0.75);\draw[dashed] (0,0.75)--(1.299,0.75); %[dashed]表示虚线 
	   %画点
	   \fill (30:1.5cm) circle(2pt);\node[right]at (30:1.5cm) {P};
	   \node[below] at (1.299,0){$\cos\theta$};\node[left] at (0,0.75){$\sin\theta$}; 
	   \node[above] at (30:2.5){$r=1$};
		%画弧
	   \pic["$\theta $",draw=black!100,angle eccentricity=1.4, angle radius=0.5cm] %感叹号后面指不透明度 需要箭头加上<->
	   {angle=c--o--a};%\alpha的位置由eccentricity决定. 
   \end{tikzpicture}
   \column{.45\textwidth} % Left column and width
   Let $P(x,y),$then\[x=\cos\theta,y=\sin\theta.\]

   For the definition of radian \[\theta=\frac{l}{r}=l,\]where $l$ is arc length of the diagonal Angle of $\theta$.

   Hence, The fan area \[S=\frac{1}{2}\theta r^2=\frac{1}{2}\theta,\]such that $\jcs{\theta=2S}.$
\end{columns} 
\end{frame}

\section{The definition of hyperbolic function}
\begin{frame}{The definition of hyperbolic function}
	\begin{columns}[c]
		\column{.45\textwidth} % Left column and width
	Likewise, for the unit hyperbola,
	\includegraphics[width=5.75cm]{sqx2.png}
	San POB is called \jc{hyperolic San}
	\column{.45\textwidth} % Left column and width
	\begin{definition}

		If the area of hyperbolic san $\overset{\frown}{PBC}$ satisfies $\jcs{S=\frac{\theta}{2}}$ , then we consider $\theta$ as \jc{hyperbolic angle}.
		Let $P(x,y)$ , define
		\[\begin{cases} x=\cosh \theta \\ y=\sinh \theta
		\end{cases}\]
		Substituting $x,y$ into function, we obtain\[\cosh^2\theta-\sinh^2\theta=1.\]
		%注:$h$是$hyperbolic$(双曲)的缩写.
	\end{definition}
\end{columns} 
\end{frame}

\section{Derivation the formula}
\begin{frame}{Derivation the formula}
		Now let begin our derivation. It is a bit tedious!
		Frow the picture, we know $x \geqslant 1$. W.L.O.G, assume $y \geqslant 0$, by equation $x^2-y^2=1,$ we obtain $y=\sqrt{x^2-1}$ such that
		\begin{align*}
			S&=S_{\vartriangle OPC}-S_{\overset{\frown}{PBC}}\\&=\frac{1}{2}xy-\int_{1}^{x} \sqrt{x^2-1} \,\mathrm{d}x. 
		\end{align*}
		Hence,\[\theta=2S=xy-2\int_{1}^{x} \sqrt{x^2-1}\,\mathrm{d}x.\]
		So the key is to compute indefinite integral
		\[\int \sqrt{x^2-1}\,\mathrm{d}x.\]
\end{frame}

\begin{frame}{Derivation the formula}
	Set $x=\sec\theta,$ then $dx=\sec\theta\tan\theta d\theta,$
	
	by $\tan^2\theta+1=\sec^2\theta$, we get $\sqrt{x^2-1}=\tan\theta$, such that
    \begin{align*}
        \int \sqrt{x^2-1}\,\mathrm{d}x&=\int \tan^2\theta \sec\theta\,\mathrm{d}\theta\\
                                  &=\int (\sec^2\theta-1)\sec\theta \,\mathrm{d}\theta\\
                                  &=\int \sec^3\theta\,\mathrm{d}\theta-\int \sec\theta\,\mathrm{d}\theta.
    \end{align*}
	Let's do it separately.
\end{frame}

\begin{frame}{Derivation the formula}
	Firstly, we deal with $\int \sec\theta\,\mathrm{d}\theta$ which needs a bit skills.
    \begin{align*}
        \int \sec\theta\,\mathrm{d}\theta &=\int \sec\theta \times \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta} \,\mathrm{d}\theta\\
                                 &=\int \frac{\sec^{2}\theta+\sec\theta\tan\theta}{\sec\theta+\tan\theta} \,\mathrm{d}\theta\\
                                &=\ln |\sec\theta+\tan\theta|+C.
    \end{align*}
	The final result comes from discovering that the numerator is exactly the derivative of the denominator.
\end{frame}

\begin{frame}{Derivation the formula}
	Next, we think about $\int \sec^3\theta\,\mathrm{d}\theta.$ Extract $\sec^2\theta$, like this
    \[\int \sec^3\theta\,\mathrm{d}\theta=\int \sec^2\theta\sec\theta\,\mathrm{d}\theta.\]
	Using integration by parts, let $u=\sec\theta,dv=\sec^2\theta\,\mathrm{d}\theta$. Taking the derivative of u and integrating dv, we get
    \[du=\sec\theta\tan\theta\,\mathrm{d}\theta,\,v=\tan\theta.\]
    Now by integration by parts, we obtain
    \[\int u\,dv=uv-\int v\,du\]
    \[\int \sec^2\theta\sec\theta\,\mathrm{d}\theta=\sec\theta\tan\theta-\int \sec\theta\tan^2\theta\,\mathrm{d}\theta.\]
\end{frame}

\begin{frame}{Derivation the formula}
	Using $\tan^2\theta+1=\sec^2\theta$ again, the integral on the right hand side of this equation can be written as
	\begin{align*}
        \int \sec\theta\tan^2\theta\,\mathrm{d}\theta&=\int \sec\theta(\sec^2\theta-1)\,\mathrm{d}\theta\\
                                            &=\jcs{\int \sec^3\theta\,\mathrm{d}\theta}-\int \sec\theta\,\mathrm{d}\theta.
    \end{align*}
	Step by step back substitution and transposition simplification, we obtain
	\[\int \sec^3\theta\,\mathrm{d}\theta=\frac{1}{2}(\sec\theta\tan\theta+\int \sec\theta\,\mathrm{d}\theta).\]
	Hence,\[\int(\sec^3\theta-\sec\theta)\,\mathrm{d}\theta=\frac{1}{2}(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|)+C.\]
\end{frame}

\begin{frame}{Derivation the formula}
	In terms of original variable, we obtain
	\[\int \sqrt{x^2-1}\,\mathrm{d}x=\frac{1}{2}(x\sqrt{x^2-1}-\ln|x+\sqrt{x^2-1}|)+C.\]
    For $Newton-Leibniz$ formula,
    \[\int_{1}^{x} \sqrt{x^2-1} \,\mathrm{d}x= \frac{1}{2}(x\sqrt{x^2-1}-\ln|x+\sqrt{x^2-1}\,|).\]
	Substituting into the original expression of $\theta$ and by $y=\sqrt{x^2-1}$, we can get
	\begin{align*}
        \theta &=xy-x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|\\
               &=xy-xy+\ln|x+y|\\
               &=\ln(x+y).
    \end{align*}
\end{frame}

\begin{frame}{Derivation the formula}
	Hence,\[x+y=e^{\theta}.\]
	Since $x$ and $y$ satisfy $x^2-y^2=1$, such that
    \[x-y=\frac{x^2-y^2}{x+y}=e^{-\theta}.\]
    Through the above two equations, we get
    \[x=\frac{e^{\theta}+e^{-\theta}}{2},y=\frac{e^{\theta}-e^{-\theta}}{2}.\]
    By the definition of hyperbolic function $x=\cosh\theta, y=\sinh\theta$, we finally obtain
    \[\begin{cases}
        \cosh\theta=\frac{1}{2}(e^{\theta}+e^{-\theta}) \\
        \sinh\theta=\frac{1}{2}(e^{\theta}-e^{-\theta}).
    \end{cases}\]
\end{frame}

\section{Functions of Complex Variables}
\begin{frame}{Expression}
	For any $z\in \mathbb{C}$,
	\[\begin{cases}
		\cos z=\frac{1}{2}(e^{iz}+e^{-iz}), \\
		\sin z=\frac{1}{2i}(e^{iz}-e^{-iz}).
	\end{cases}\]
	\[\begin{cases}
        \cosh z=\frac{1}{2}(e^{z}+e^{z}),\\
        \sinh z=\frac{1}{2}(e^{z}-e^{z}).
    \end{cases}\]
	Actually,\[\sinh iz=i\sin z.\]
	Multiply $i$ just means rotating, hence, \jc{their pictures in the complex number domain are essentially the same.}
\end{frame}

\section{Reference material}
\begin{frame}{Reference Website Link}
\begin{compactenum}
    \item https://www.bilibili.com/video/BV1xp4y1v7cw 
	
	双曲函数——带你领略课本上没有的神奇函数！ 
    \item https://www.bilibili.com/video/BV1hp4y1v7AR 
	
	三角函数与双曲函数的奇妙联系——浅谈复变三角与双曲
    \item https://zhuanlan.zhihu.com/p/20042215
	
	可能是最好的讲解双曲函数的文章-王希的文章-知乎
	\item https://www.latexstudio.net/index/details/index?mid=993
	
	华中师范大学数学与统计学学院主题 Beamer
\end{compactenum}
\end{frame}

\begin{frame}{Bibliography}
\nocite{*}
\bibliography{reference}
\end{frame}

\begin{frame}
\begin{center}
{\Huge\calligra Thanks!}
\end{center}
\end{frame}

\end{document} 